Proof that Markov Property is not Satisfied at any Order? My textbook has this figure in it:

The textbook then says, Using d-separation, we can see there is always a path connecting $x_n$ and $x_{m}$ via the latent variables. This makes sense to me because if I use the rules of d-separation, I can tell that no path between any of the $x$ variables is blocked by any of the latent $z$ variables (because they aren't observed) and it's like that tail-to-tail thing.
What I don't understand is this, "Thus the predictive distribution $p(x_{n+1} | x_1,...,x_n)$ for observation $x_{n+1}$ given all previous observations does not exhibit any conditional independence properties, and so our predictions for $x_{n+1}$ depend on all previous observations. The observed variables, however, do not satisfy the Markov property at any order." Aren't the "observed variables" just the $x_i$s here, so how is this sentence any different from the previous? It almost sounds like the observed variables are something else?
I think what this is saying (but I could be misunderstanding it) is that for any Markov order (like 3rd order model where observation depends on previous 3 observations) the Markov property will always be violated because $x_{n+1}$ always depends on all previous $x_n$. However, this sentence seems to be wrong because if we are predicting $x_{n+1}$ this does in fact satisfy the Markov property at order $n$ (i.e. a Markov model that depends on the previous $n$ observations). So why isn't this sentence wrong? Or am I misunderstanding it? 
 A: For the Markov property (of order $k$) to hold for the sequence $x_1,x_2,\ldots,$ the conditional distribution of $x_n$ given the previous values $x_1\ldots,x_{n-1}$ should be equal to the conditional distribution conditioning only on $x_{n-k},x_{n-k+1},\ldots,x_{n-1}$, that is,
\begin{equation}
\forall n\geq k: p(x_n \mid x_1,\ldots,x_{n-1}) = p(x_n \mid x_{n-k}, \ldots, x_{n-1})
\end{equation}
(The $n\geq k$ is there because for $n<k$ there is not enough previous values, so the condition is trivially statisfied).
Intuitively, (in terms of predicting), the process forgets its history except the $k$ previous values. This does not occur for the given process, since the conditional distribution always depends on the total history. 
In the question, you point out that indeed if $n=k$ (where $k$ is the Markov order), the property is satisfied (which is obvious since the process cannot remember its earlier history since the earlier history does not even exist). This observation is trivial in the sense that it would hold for any joint distribution $x_1,\ldots,x_n$, and does not tell anything interesting about the process. The process, however, does not have Markov property of order $n$ since, say, 
\begin{equation}
p(x_{n+607} \mid x_1,\ldots,x_{n+606}) \neq p(x_{n+607} \mid x_{607}, \ldots, x_{n+606}).
\end{equation}
Caveat: Actually we should only say that the Markov property is not implied by the graphical model - it's always possible to have more conditional independencies than implied by the graphical model structure. If, for example, the $x$s are independent of the $z$s (and everything), the sequence is obviously Markovian but still satisfies all conditional independence properties described by the graphical model. (In the same sense as in a graphical model consisting of two nodes and one edge, the nodes may still happen to be independent).
A: I think that sentence is just to show the difference between sequence $x$ and $z$ in terms of conditional independencies, which is, as you probably understand already, the latent variables satisfy the first order Markov property, while the observed variables do not.
If we assume an n-order Markov property, we'll have something like
$$x_{n+1}\perp\!\!\!\perp\emptyset \mid\{x_n, x_{n-1},...,x_0\},$$
which says $x_{n+1}$ is independent of nothing given the observations, which basically means the same as said in the previous sentence that "our predictions for $x_{n+1}$ depend on all previous observations". 
